$f\,^{\prime}(x)=\dfrac{81}{x^4}$ and $f(-1)=18$. $f(3) = $
Solution: Finding $f(x)$ We have $f'(x)=\dfrac{81}{x^4}$ and we want to find $f(x)$ : $\begin{aligned}f(x) &= \int f'(x)\,dx \\\\ & = \int \dfrac{81}{x^4}\,dx \\\\ & = {-27x^{-3}} {+ C} \end{aligned}$ Finding $ C$ Goal: We need to find $ C$ such that $f(-1)=18$. Here's what we get when we plug in $-1$ : $\begin{aligned}f(-1)&={-27(-1)^{-3}} {+ C}\\\\ &={27} {+ C} \end{aligned}$ We are given that this must equal $18$ : $18 = {27} {+ C}$ Solving the equation gives us ${C=-9}$. Finding $f(3)$ Now, we have that $f(x)= {-27x^{-3}} {-9}$. Let's find $f(3)$ by plugging in $3$ : $\begin{aligned}f(3)&=-27(3)^{-3}-9\\\\ &=-10 \end{aligned}$ The answer $f(3) = -10$